/*
	d[i][j] = number of ways to produce i with coins from c[1] to c[j]
	d[i][j] sum of d[i-k*c[j]][j-1] where k = 1 to where possible
*/

#include <iostream>

using namespace std;

int c[6] = {0, 1, 5, 10, 25, 50};
long long d[30001][6];
int n;

int main()
{	
	for (int j = 0; j < 6; j++) d[0][j] = 1;
	
	for (int i = 1; i <= 30000; i++)
	{
		for (int j = 1; j < 6; j++)
		{
			d[i][j] = d[i][j-1];
			int k = 1;
			while (i - k*c[j] >= 0)
			{
				d[i][j] += d[i - k*c[j]][j-1];
				k++;
			}
		}
	}
	
	
	while (cin >> n)
	{
		if ((n == 0) || (d[n][5] == 1))
			cout << "There is only 1 way to produce " << n << " cents change." << endl;
		else
			cout << "There are " << d[n][5] << " ways to produce " << n << " cents change." << endl;
	}
	
	return 0;
}
